% statex.m - state-space example (done in class) % 332:345 - Linear Systems & Signals - Fall 2009 clear all; syms t u s real syms y0 y1 real % notation: y0 = y(0-) and y1 = dy/dt(0-) f = exp(-u) % input num = [2 13 8]; den = [1 5 6]; % system [A,B,C,D] = tf2ss(num,den) % convert to state-space PHI = inv(s*eye(2) - A) % Laplace transform of state-transition matrix phi = expm(A*t) % state transition matrix h = C*expm(A*t)*B % impulse response h(t), must add to it the D*delta(t) term F = sym([C; C*A]) % observability matrix for an order-2 system x0= inv(F)*[y0;y1] % map initial conditions [y0; y1] to initial state vector x(0-) xf = simplify(int(expm(A*(t-u))*B*f, u, 0,t)) % forced part of the state xh = simplify(expm(A*t)*x0) % homogeneous part due to initial state x = simplify(xh+xf) % full state vector y = simplify(C*x + D*subs(f,u,t)) % output obtained from state eq yf = simplify(int(subs(h,t,t-u)*f,u,0,t) + D*subs(f,u,t)) % forced output via convolution with h(t) yh = simplify(C*expm(A*t)*x0) % homogeneous output simplify(y-yh-yf) % test equality of y and yh+yf % -------- diagonal realization ------------ [r,p,k] = residue(num,den) % verify the PFE coefficients A = diag([-3,-2]); B = [1;1]; C = [13,-10]; D = 2; % state-space parameters phi = expm(A*t) % diagonal transition matrix h1 = C*phi*B; % impulse response simplify(h-h1) % should be the same as above F = sym([C; C*A]); % observability matrix invF = inv(F) x0= inv(F)*[y0;y1] % map initial conditions to x(0) xf = simplify(int(expm(A*(t-u))*B*f, u, 0,t)) % forced part of the state xh = simplify(expm(A*t)*x0) % homogeneous part due to initial state x = simplify(xh+xf) % full state vector y1 = simplify(C*x + D*subs(f,u,t)) % output y(t) yf = simplify(int(subs(h,t,t-u)*f,u,0,t) + D*subs(f,u,t)) % forced output yh = simplify(C*expm(A*t)*x0) % homogeneous output simplify(y-y1) % verify y(t) is the same
f =
exp(-u)
A =
-5 -6
1 0
B =
1
0
C =
3 -4
D =
2
PHI =
[ s 6 ]
[------------ - ------------]
[ 2 2 ]
[s + 5 s + 6 s + 5 s + 6]
[ ]
[ 1 s + 5 ]
[------------ ------------ ]
[ 2 2 ]
[s + 5 s + 6 s + 5 s + 6 ]
phi =
[-2 exp(-2 t) + 3 exp(-3 t) 6 exp(-3 t) - 6 exp(-2 t)]
[ ]
[ -exp(-3 t) + exp(-2 t) 3 exp(-2 t) - 2 exp(-3 t)]
h =
-10 exp(-2 t) + 13 exp(-3 t)
F =
[ 3 -4]
[ ]
[-19 -18]
x0 =
[ 9/65 y0 - 2/65 y1 ]
[ ]
[ 19 ]
[- --- y0 - 3/130 y1]
[ 130 ]
xf =
[- 1/2 (-4 exp(t) + 3 + exp(2 t)) exp(-3 t)]
[ ]
[ 1/2 (1 - 2 exp(t) + exp(2 t)) exp(-3 t) ]
xh =
[3/5 exp(-2 t) y0 + 1/5 exp(-2 t) y1 - 6/13 exp(-3 t) y0
- 3/13 exp(-3 t) y1]
[2/13 exp(-3 t) y0 + 1/13 exp(-3 t) y1 - 3/10 exp(-2 t) y0
- 1/10 exp(-2 t) y1]
x =
[3/5 exp(-2 t) y0 + 1/5 exp(-2 t) y1 - 6/13 exp(-3 t) y0
- 3/13 exp(-3 t) y1 + 2 exp(-2 t) - 3/2 exp(-3 t) - 1/2 exp(-t)]
[2/13 exp(-3 t) y0 + 1/13 exp(-3 t) y1 - 3/10 exp(-2 t) y0
- 1/10 exp(-2 t) y1 + 1/2 exp(-3 t) - exp(-2 t) + 1/2 exp(-t)]
y =
3 exp(-2 t) y0 + exp(-2 t) y1 - 2 exp(-3 t) y0 - exp(-3 t) y1 + 10 exp(-2 t)
- 13/2 exp(-3 t) - 3/2 exp(-t)
yf =
10 exp(-2 t) - 13/2 exp(-3 t) - 3/2 exp(-t)
yh =
3 exp(-2 t) y0 - 2 exp(-3 t) y0 + exp(-2 t) y1 - exp(-3 t) y1
ans =
0
r =
13.0000
-10.0000
p =
-3.0000
-2.0000
k =
2
phi =
[exp(-3 t) 0 ]
[ ]
[ 0 exp(-2 t)]
ans =
0
invF =
[-2 -1]
[-- --]
[13 13]
[ ]
[-3 -1]
[-- --]
[10 10]
x0 =
[- 2/13 y0 - 1/13 y1]
[ ]
[- 3/10 y0 - 1/10 y1]
xf =
[- 1/2 exp(-3 t) + 1/2 exp(-t)]
[ ]
[ -exp(-2 t) + exp(-t) ]
xh =
[- 1/13 exp(-3 t) (2 y0 + y1)]
[ ]
[- 1/10 exp(-2 t) (3 y0 + y1)]
x =
[- 2/13 exp(-3 t) y0 - 1/13 exp(-3 t) y1 - 1/2 exp(-3 t) + 1/2 exp(-t)
]
[- 3/10 exp(-2 t) y0 - 1/10 exp(-2 t) y1 - exp(-2 t) + exp(-t)]
y1 =
3 exp(-2 t) y0 + exp(-2 t) y1 - 2 exp(-3 t) y0 - exp(-3 t) y1 + 10 exp(-2 t)
- 13/2 exp(-3 t) - 3/2 exp(-t)
yf =
10 exp(-2 t) - 13/2 exp(-3 t) - 3/2 exp(-t)
yh =
3 exp(-2 t) y0 - 2 exp(-3 t) y0 + exp(-2 t) y1 - exp(-3 t) y1
ans =
0